设Sn是等差数列{an}的前n项和,S7=3(a2+a12),则a7a4的值为( ) A.16 B.13 C.35 D.76
问题描述:
设Sn是等差数列{an}的前n项和,S7=3(a2+a12),则
的值为( )a7 a4
A.
1 6
B.
1 3
C.
3 5
D.
7 6
答
由S7=3(a2+a12)结合等差数列的求和公式可得
=3(a2+a12),7(a1+a7) 2
再由等差数列的性质可得
=3×2a7,即7a4=6a7,7×2a4
2
变形可得
=a7 a4
7 6
故选D